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Physics
OpenStudy (anonymous):

What is current a measure of in the photoelectric effect?

OpenStudy (anonymous):

How much is the current produced by the photoelectric effect? Is that the question? I ask for me and for anyone desiring to answer you.

OpenStudy (anonymous):

I am more interested in whether it is an accurate measurement of the amount of photons incident on the surface of the metal.

OpenStudy (anonymous):

Photoelectric sensors are not 100% efficient, their capacity as generators depends of the purity and conformation of the receiver surface (so far silicon). You would need a circuit to multiply the voltage produced and then also use a photometer to create a graph involving both photons and electrons. I expect changes in the graph even for wafers of the same lot produced. It would be more accurate to measure photons with a photo-sensor similar to those in cameras.

OpenStudy (anonymous):

If you're interested in the practice, getting the percentage efficiency and voltage output can roughly give you the photon's account. Yes, why not.

OpenStudy (anonymous):

the current is proportional to the intensity of incident light. so you can say it's a measure of intensity of the incident light.

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